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Chapter 2: Problem 19

In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) in \(1.5 \mathrm{~s}\). Assuming that the player accelerates uniformly, determine the distance he runs.

Short Answer

Expert verified
The distance the player runs is 4.5 meters.

Step by step solution

01

Identify Given Information

The problem tells us that the player starts from rest, which means the initial velocity (\(u\)) is \(0 \text{ m/s}\), the final velocity (\(v\)) is \(6.0 \text{ m/s}\), and the time (\(t\)) taken is \(1.5 \text{ s}\). The player's acceleration is uniform.
02

Use the Formula for Uniform Acceleration

To find the distance covered, we can use the kinematic equation that relates distance (\(s\)), initial velocity (\(u\)), time (\(t\)), and acceleration (\(a\)), which is \(s = ut + \frac{1}{2}at^2\). We don't know \(a\) yet, so we'll find it first.
03

Calculate Acceleration

Since we have the initial velocity, final velocity, and time, we can find acceleration using the formula \(a = \frac{v - u}{t}\). Substituting the known values gives \[a = \frac{6.0 \text{ m/s} - 0 \text{ m/s}}{1.5 \text{ s}} = 4.0 \text{ m/s}^2\].
04

Calculate the Distance

Now, substitute the values of \(u = 0\), \(t = 1.5 \text{ s}\), and \(a = 4.0 \text{ m/s}^2\) into the equation \(s = ut + \frac{1}{2}at^2\). Calculate: \[s = 0 \times 1.5 + \frac{1}{2} \times 4.0 \times (1.5)^2\] \[s = 0 + 0.5 \times 4.0 \times 2.25\] \[s = 4.5 \text{ meters}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate over time. This means that for every second, the object gains or loses the same amount of velocity. In our basketball player example, the player accelerates uniformly from rest up to 6.0 m/s. This consistent change in velocity allows us to use kinematic equations to calculate various aspects of motion, such as distance and acceleration. These equations are powerful tools in classical mechanics, as they help in predicting future positions and velocities of moving objects.
Initial Velocity
The initial velocity is the speed at which an object begins its motion. In any problem involving motion, identifying the initial velocity is crucial as it sets the starting point for the calculations. In the basketball player's case, since he starts from rest, his initial velocity is 0 m/s. This simplifies our calculations considerably. Often, in problems with initial velocity being zero, the equations tend to reveal the role of external forces, such as acceleration, more clearly.
Final Velocity
Final velocity refers to the speed of an object at the end of the observed time interval. This value can be directly provided or derived using kinematic equations. For the basketball player, his final velocity is given as 6.0 m/s. Understanding how final velocity is reached by uniform acceleration helps us bridge the relationship between different states of motion over time. It also highlights the impact of constant acceleration on an object's speed, allowing us to trace slow starts or rapid speeds depending on context.
Distance Calculation
Distance calculation in uniformly accelerated motion can be carried out using the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \] where \(s\) is the distance, \(u\) is the initial velocity, \(t\) is time, and \(a\) is acceleration. Since we know the initial velocity is 0 m/s and the acceleration is 4.0 m/s虏, we can compute how far the player runs in 1.5 seconds. Substituting these values into the equation gives us 4.5 meters. Distance calculation not only tells us how far an object has moved but also provides a spatial context to the motion described by the other kinematic parameters.

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Most popular questions from this chapter

A locomotive is accelerating at \(1.6 \mathrm{~m} / \mathrm{s}^{2}\). It passes through a 20.0 -m-wide crossing in a time of 2.4 s. After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{~m} / \mathrm{s} ?\)

A ball is thrown upward from the top of a 25.0 -m-tall building. The ball's initial speed is \(12.0 \mathrm{~m} / \mathrm{s}\). At the same instant, a person is running on the ground at a distance of \(31.0 \mathrm{~m}\) from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). The car is a distance \(d\) away. The bear is \(26 \mathrm{~m}\) behind the tourist and running at \(6.0 \mathrm{~m} / \mathrm{s}\). The tourist reaches the car safely. What is the maximum possible value for \(d ?\)

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The stones are thrown with the same speed. (a) Does the stone thrown upward gain or lose speed as it moves upward? Why? (b) Does the stone thrown downward gain or lose speed as time passes? Explain. (c) The speed at which the stones are thrown is such that they cross paths. Where do they cross paths, above, at, or below the point that corresponds to half the height of the cliff? Justify your answer. Problem The height of the cliff is \(6.00 \mathrm{~m},\) and the speed with which the stones are thrown is \(9.00 \mathrm{~m} / \mathrm{s}\). Find the location of the crossing point. Check to see that your answer is consistent with your answers to the Concept Questions.

In 1998 , NASA launched Deep Space \(I\) (DS-1), a spacecraft that successfully flew by the asteroid named 1992 KD (which orbits the sun millions of miles from the earth). The propulsion system of DS-1 worked by ejecting high-speed argon ions out the rear of the engine. The engine slowly increased the velocity of DS-1 by about \(+9.0 \mathrm{~m} / \mathrm{s}\) per day. (a) How much time (in days) would it take to increase the velocity of DS-1 by \(+2700 \mathrm{~m} / \mathrm{s}\) ? (b) What was the acceleration of \(\mathrm{DS}-1\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right) ?\)
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